c - 当您将整数值类型转换为 char 指针时会发生什么？

Question

例如，

char * integerToString(void);

int main() {
    char *myString;
    do {
        myString = integerToString();
    } while (myString == (char *)-1); // worked as intended
    free(myString);
    return 0;
}

char * integerToString(void) {

    int userInput;
    printf("Enter an integer: ");
    scanf("%d", &userInput);

    if (userInput < 0 || userInput > 99)
        return (char *)-1; // what happens here?

    char *myString = (char *)malloc(sizeof(char) * 2);
    myString[0] = (int)floor(userInput/10.0) + '0';
    myString[1] = userInput%10 + '0';
    return myString;
}

并且程序按预期工作，但是当您将整数值(没有将整数分配给变量)键入字符指针时，究竟会发生什么？这个程序会一直有效吗？谢谢。

Answer 1

C99:

6.3.2.3 Pointers

3. An integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant. If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function.

[...]

5. An integer may be converted to any pointer type. Except as previously specified, the result is implementation-defined, might not be correctly aligned, might not point to an entity of the referenced type, and might be a trap representation.

因此将-1 转换为指针具有实现定义的结果。因此答案是否定的:这不能保证在一般情况下有效。

---

特别是:如果它确实是一个陷阱表示，您的代码将与以下方面发生冲突:

6.2.6 Representation of types

6.2.6.1 General

[...]

5. Certain object representations need not represent a value of the object type. If the stored value of an object has such a representation and is read by an lvalue expression that does not have character type, the behavior is undefined. If such a representation is produced by a side effect that modifies all or any part of the object by an lvalue expression that does not have character type, the behavior is undefined. Such a representation is called a trap representation.

即while (myString == (char *)-1); 如果 myString 是陷阱表示，则有未定义的行为。